Determining wattage - Please explain volts

[ChrisIsBored]

Hey guys,

I'm in need of a way to calculate the approximate wattage an appliance uses from the given Amps listed that the product uses.

I'm aware that Watts = Volts X Amps, but I don't know much about volts. I've now seen two sites with one giving 110 volts and another giving 115 volts for finding out how much an appliance uses. What's the difference and the standard for a US household?

If anyone is interested, i'm creating a program that calculates how much your electric bill will cost using [X] number of watts for aquaria related products... but there are a few products which are listed in Amps rather than Watts.

 

[Mark R]

I don't know what the nominal voltage in the US is, but it's going to make no real difference to your calculation if you use 110 or 115 V - this is only a 5% inaccuracy, there are some other far more significant sources of inaccuracy:

1) heaters (which will consume the most power) aren't on all the time, they are usually thermostatically controlled - average power use will therefore depend on size of tank, and temperature of the room, etc.

2) Where the manufacturer states 'amps' or 'watts' on a device, this is usually the maximum it will take (rather than what it usually takes) - you may over estimate substantially with this.

3) You need to take into account things on timers, and how much they are on per day(e.g. lights).

4) If the device is a fluorescent light or motor (pump), then 'Watts = Volts x Amps' doesn't work.

 

[Fandu]

Normal household voltages in north america are 110-120V RMS. So 115 is an average of an average, it's just the easiest to calculate with.

BTW - be aware that some hydro companies bill per VA, not watts. However, understanding VA requires a working knowledge of electionics, calculus, and imaginary power. But basically, every circuit (ie your house) has either capacitive of inductive properties, so it stores energy in the circuit. However storing power, also requires power (imaginary power), and VA is a measure of the real power (watts), plus imaginary power. So to calculate the VA draw of an applicance, you have to know it's power factor (pf). Some applicanes will give the pf value, and some don't.

 

[ChrisIsBored]

Thanks guys...

>>4) If the device is a fluorescent light or motor (pump), then 'Watts = Volts x Amps' doesn't work<<

Could you explain this statement Mark?

And since this is really for salt water aquaria... the lights are overwhelmingly the highest costing items. I've taken into account the timers and such... in fact you can change the option of how many hours per day an appliance actually runs. But in this hobby, there are a lot of things like pumps that only give us a amp draw rather than watts.

>>2) Where the manufacturer states 'amps' or 'watts' on a device, this is usually the maximum it will take (rather than what it usually takes) - you may over estimate substantially with this.<<

So my 110W VHO bulb will only reach 110W if at 100% efficiency then correct? I guess it would depend on the ballast of the light, but would the wattage fluctuate? I mean after being on for a few hours, would it continuously pull the same amount of power?

Here's the work... calculator

 

[SharkyTM]

well, as far as i know, light-watts, like a 50 watt bulb, is the amount of light energy given off... not the power watts used. Therefore, it may actually be a 100 watt draw- 100v, 1 Amp, but may only be a 50 watt bulb. 90% of the energy used is converted to heat, not light...

To understand amps and volts, consider this- the amperage is actually a measure of the number of electrons passing by a particular point. An ampere is 1 coloumb/second. 1 coulomb is a mole of electrons. 6.022x10^23 things in a mole.
Voltage is the "pressure" behind the electrons. High voltage will travel farther thru a wire than a lower voltage...

wattage is the overall amount of power used.

Hope this helps,
SharkyTM

 

[SharkyTM]

check this out...
Electricity

 

[dejitaru]

The standard for USA is 120 volts.

'X' amps * 120VAC
Most sockets I've tested were about 125V. A 110V socket could power a 125V device, or the reverse. There is a great margin of error when working with volts.

assume 120

 

[Mark R]

>>4) If the device is a fluorescent light or motor (pump), then 'Watts = Volts x Amps' doesn't work<<

Could you explain this statement Mark?

It's to do with 'power factor' a feature of AC circuits which contain an inductor (a coil of wire). Motors contain wire coils (that's how they work), and a inductor is used to limit the current flow in a fluorescent light.

It means that some of the time, the current (amps) taken by the device operates against zero volts (or even negative volts). So the device actually takes more amps than you would expect for the number of watts it uses.

I have a compact fluorescent light bulb lying on my desk at the moment - it says 21 W, 230 V, 170 mA. Do the sums yourself, to see what I mean.

Where I am, for home use, the electricity companies must bill you only for Watts. Elsewhere, some charge you for Volt-amps (volts x amps).

well, as far as i know, light-watts, like a 50 watt bulb, is the amount of light energy given off

Nope - A 50 W light bulb takes 50 W from the mains, a 100 W light bulb takes 100 W from the mains, of which about 2W is converted to light (for a filament bulb). Be careful with fluorescent lamps, the ballast often takes extra power (e.g. the tube may be 50W but the ballast might use another 10).

So my 110W VHO bulb will only reach 110W if at 100% efficiency then correct

For bulbs, this isn't much of a problem - I was thinking more about devices with multiple settings (e.g. heaters). Bulbs don't significantly vary in power consumption with time.

 

[PowerEngineer]

Mark's got it right.

In DC systems, the watts are in fact equal to the current times the voltage. And that's actually true in AC systems too, but the difference is that both the voltage and current are 60 Hz sinusoidal waves, and the two waves are usually out of phase (i.e. their peaks to not occur at the same instants in time). So the power being delivered at any instant is the voltage at that instant times the voltage at that instant, meaning that the power delivery also changes from instant to instant. But we want to simplify this if we can, so we start talking about the average power delivered each cycle. This leads into the determination of so-call RMS values (RMS stands for root-mean-square) which are the peak values (for voltage and current) divided by the square root 0f 2. The real power delivered is the RMS value of the voltage multiplied by the RMS value of the current times the cosine of the angle between the voltage and the current.

And all customer electric meters that I've ever seen are designed to measure real power consumption only.

 

[Krakerjak]

Originally posted by: Mark R

>>4) If the device is a fluorescent light or motor (pump), then 'Watts = Volts x Amps' doesn't work<<

Could you explain this statement Mark?

It's to do with 'power factor' a feature of AC circuits which contain an inductor (a coil of wire). Motors contain wire coils (that's how they work), and a inductor is used to limit the current flow in a fluorescent light.

It means that some of the time, the current (amps) taken by the device operates against zero volts (or even negative volts). So the device actually takes more amps than you would expect for the number of watts it uses.

I have a compact fluorescent light bulb lying on my desk at the moment - it says 21 W, 230 V, 170 mA. Do the sums yourself, to see what I mean.

Where I am, for home use, the electricity companies must bill you only for Watts. Elsewhere, some charge you for Volt-amps (volts x amps).

well, as far as i know, light-watts, like a 50 watt bulb, is the amount of light energy given off

Nope - A 50 W light bulb takes 50 W from the mains, a 100 W light bulb takes 100 W from the mains, of which about 2W is converted to light (for a filament bulb). Be careful with fluorescent lamps, the ballast often takes extra power (e.g. the tube may be 50W but the ballast might use another 10).

So my 110W VHO bulb will only reach 110W if at 100% efficiency then correct

For bulbs, this isn't much of a problem - I was thinking more about devices with multiple settings (e.g. heaters). Bulbs don't significantly vary in power consumption with time.

I was under the impression that the MAJORITY of loads in a residential home are resistive, which is why power companies only charge for real power and ignore the reactive power draw from homes.
Hence single phase outlets in homes b/c neutral really is neutral since no reactive power is returning to the supply.

 

[PowerEngineer]

The "power factor" is the cosine of the phase angle between the voltage and current waves.

Many home loads are esssentially pure resistance, such as incandesent light bulbs and any heating elements (e.g. toasters, water heaters, and base-board heating). There are many other loads that have some reactive components; anything with a transformer in it (like floresent bulbs and all the power supplies in your electronic equipment) is somewhat reactive; anything with a motor in it (like your refrigerator, freezer, wash machine, and heat pumps) are also somewhat reactive. That said, the power factors on these devices are normally greater than 0.8, so the overall characteristic of a normal home is dominantly resistive.

Do not assume that there is no current flowing through the neutral wire!! While it is true that a balanced three phase electrical system serving a balanced three phase load will have zero current at the neutral, the vast majority of homes receive only single-phase service (even the 120 V and 240 V circuits are from the same electrical phase). Only homes with larger motor loads that are more effecieint with multiple phases get another phase delivered to them on another wire. This means that current flowing into a load from the hot side in your home is definitely flowing out of the load into the neutral wire.